sgu194 Reactor Cooling【无源汇有上下界可行流】

　　这是模板题了吧，先建立附加源汇，然后保留每个点的in-out，如果这个值是正的，那么就从附加源先这个点连一个边权为in-out的边，否则从这个点向附加汇连一条相反数的边，剩下题目中的边就用上界-下界连就好了。

#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define drep(i, a, b) for (int i = a; i >= b; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define mp make_pair
#define pb push_back
#define clr(x) memset(x, 0, sizeof(x))
#define xx first
#define yy second
using namespace std;
typedef long long i64;
typedef pair<int, int> pii;
const int inf = ~0U >> 1;
const i64 INF = ~0ULL >> 1;
//***********************************

const int maxn = 205, maxm = 40205;

struct Ed {
    int u, v, nx, c; Ed() {}
    Ed(int _u, int _v, int _nx, int _c) :
        u(_u), v(_v), nx(_nx), c(_c) {}
} E[maxm << 1];
int G[maxn], edtot = 1;
void addedge(int u, int v, int c) {
    E[++edtot] = Ed(u, v, G[u], c);
    G[u] = edtot;
    E[++edtot] = Ed(v, u, G[v], 0);
    G[v] = edtot;
}

int level[maxn], s, t;
bool bfs() {
    static int que[maxn]; int qh(0), qt(0);
    clr(level);
    level[que[++qt] = s] = 1;
    while (qh != qt) {
        int x = que[++qh]; if (qh == maxn - 1) qh = 0;
        for (int i = G[x]; i; i = E[i].nx) if (E[i].c && !level[E[i].v]) {
            level[que[++qt] = E[i].v] = level[x] + 1;
            if (qt == maxn - 1) qt = 0;
        }
    }
    return !!level[t];
}
int dfs(int u, int rm) {
    if (u == t) return rm;
    int rm1 = rm;
    for (int i = G[u]; i; i = E[i].nx) {
        if (E[i].c && level[E[i].v] == level[u] + 1) {
            int flow = dfs(E[i].v, min(E[i].c, rm));
            E[i].c -= flow, E[i ^ 1].c += flow;
            if ((rm -= flow) == 0) break;
        }
    }
    if (rm1 == rm) level[u] = 0;
    return rm1 - rm;
}

int l[maxm], in[maxn], out[maxn];
int cnt;
int main() {
    clr(G), edtot = 1, clr(in), clr(out);
    int n, m;
    scanf("%d%d", &n, &m);
    s = n + 1, t = n + 2;
    rep(i, 1, m) {
        int x, y, a, b; scanf("%d%d%d%d", &x, &y, &a, &b); l[i] = a;
        addedge(x, y, b - a);
        in[y] += a, out[x] += a;
    }
    int sum(0);
    rep(i, 1, n) {
        if (in[i] > out[i]) addedge(s, i, in[i] - out[i]), sum += in[i] - out[i];
        else addedge(i, t, out[i] - in[i]);
    }
    int ans(0);
    while (bfs()) ans += dfs(s, 0x3f3f3f3f);
    if (ans != sum) { puts("NO"); return 0; }
    puts("YES");
    for (int i = 1; i <= m; i++) printf("%d
", E[(i << 1) ^ 1].c + l[i]);
    return 0;
}